I am trying to to #3(b) from the attached image and I don't know how to go from the f'(x) = stuff part to the the "Since ln(2) > 1/2" part and the stuff that follows it. I don't get the logic there. I do understand the decreasing for x >= 2 stuff and how to deal with the inequalities algebraically but, like mentioned, I just don't get why I'm going directly from f'(x) = stuff to "Since ln(2) > 1/2".

You want to know when the function $\displaystyle \frac{x}{2^x}$ is increasing and decreasing. So, when you look at the derivative, the denominator is positive. So, all that matters is what is happening in the numerator. If it is positive, then the derivative is positive. If it is negative, then the derivative is negative. So, the question becomes, is $\displaystyle 1-x(\ln{2})$ strictly positive or strictly negative for all values $\displaystyle x\ge k$ for some $\displaystyle k \in \mathbb{R}$?

Let's find the zero for this equation:
$\displaystyle 1-x(\ln{2}) = 0$
$\displaystyle \frac{1}{\ln{2}} = x$
So, that is where the derivative is zero. When it is positive, when is it negative? So, we discover that it is negative when $\displaystyle x \ge 2$. Now, we know when the derivative is negative. So, the function is decreasing for $\displaystyle x \ge 2$. And the next part determines that the function is non-increasing.

So, if you have a function that is non-increasing, and the numerator and denominator both tend towards infinity as $\displaystyle x \to \infty$, then you can apply L'Hopital's Rule.
So, all of that "stuff" as you put it is proving that the conditions are met for applying L'Hopital's rule.